Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 2 / 1

Use $q\in k^{\times }$ and the quantum plane convention $yx=qxy$. In this convention $S{L}_q(2)$ is the coordinate Hopf algebra of quantum SL2 with the catalog parameter replaced by $q^{-1}$. It is generated by $a,b,c,d$, subject toEquivalently the last determinant relation is $da-qbc=1$. The quantum determinant convention is essential: these relations use the inverse of the parameter in the catalog's $ab=qba$ convention.

Writing $T=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$, the comultiplication is matrix multiplication, $\Delta t_{ij}={\sum }_k{t}_{ik}\otimes t_{kj}$. Explicitly,The counit is $\epsilon (a)=\epsilon (d)=1$, $\epsilon (b)=\epsilon (c)=0$. Extend $\Delta$ and $\epsilon$ as unital algebra homomorphisms over a field. Matrix multiplication is associative, giving coassociativity; multiplication by the identity matrix gives the counit laws. The antipode isFor example $(TS(T){)}_{11}=ad-q^{-1}bc=1$ and $(S(T)T{)}_{11}=da-qbc=1$, while the off-diagonal entries vanish by the commutation relations. Thus this is a Hopf algebra, not merely an algebra presentation.

Take the right coaction of the coordinate Hopf algebra of quantum SL2 on the quantum plane to beIt is extended multiplicatively and sends $1$ to $1\otimes 1$. The matrix comultiplication and counit verify the coaction identities on $x,y$. To check that it descends to the quantum plane, expandThe last bracket vanishes because $bc=cb$ and $da-ad=(q-q^{-1})bc$.

For the requested calculation, $ca=qac$ and $yx=qxy$ giveMultiply by $\delta (y)$, put all monomials into the order $x^r{y}^s$, and use $bc=cb$. The answer isA left-coaction convention transposes where the matrix coefficients occur; the displayed answer fixes the right coaction throughout.

An $R$-point means a unital algebra homomorphism over a field $\alpha :k_q[x,y]\to R$, as in points of a noncommutative algebra. It is specified by $X=\alpha (x)$ and $Y=\alpha (y)$ satisfying $YX=qXY$. For $R=\mathbb{C}$ and $q$ not a root of unity, $q\ne 1$, so scalar commutativity gives $(1-q)XY=0$. Hence the scalar points are the union of the two coordinate axes:

For matrix-valued points of a noncommutative algebra, put $H={\mathbb{C}}^n$ and let $H_{\lambda }(X)=\ker (X-\lambda I{)}^n$ be a generalized eigenspace. From $XY=q^{-1}YX$,Therefore $Y{H}_{\lambda }(X)\subseteq H_{q^{-1}\lambda }(X)$, interpreting a missing eigenvalue as the zero space. In particular the linear subspaces $V_X={⨁}_{\lambda \ne 0}{H}_{\lambda }(X)$ and $H_0(X)$ are invariant under both matrices. If $\lambda \ne 0$, the numbers $\lambda ,q^{-1}\lambda ,\dots ,q^{-n}\lambda$ are distinct. A nonzero vector $Y^{n}v$ with $v\in H_{\lambda }(X)$ would force all $n+1$ intervening generalized eigenspaces to be nonzero, impossible in dimension $n$. Thus $Y^n{V}_X=0$: $Y$ is nilpotent on $V_X$, although $X$ is invertible there.

Apply the same argument with $X,Y$ exchanged, using $XY=q^{-1}YX$. The sum $V_Y$ of the nonzero $Y$ generalized eigenspaces is invariant under both matrices, and $X$ is nilpotent on it. Consequently $V_Y\subseteq H_0(X)$. Conversely, since $Y$ is nilpotent on $V_X$, its nonzero generalized eigenspaces all lie in $H_0(X)$. Decompose the restriction of $Y$ to that invariant space into its nonzero and zero generalized eigenspaces:The zero generalized eigenspace of $Y$ is $X$-invariant by the scaling identity, so $U$ is invariant under both matrices. On $U$, both matrices have only zero eigenvalues, hence are nilpotent matrices. This proves the spectral decomposition of a quantum-plane point:Here $V_X,V_Y$ mean sums of generalized eigenspaces, not merely ordinary eigenspaces; that distinction is necessary for non-diagonalizable points.